François has excellently addressed your question 1; allow
me to address question 2. I understand the question to be:
what will be the mathematical effects if someone were to
show that there are no (weakly) inaccessible cardinals? A
similar question would apply to any of several large
cardinals. So let me list some consequences.

First, let me note that the existence of a weakly
inaccessible cardinal is provably equiconsistent with the
existence of a (strongly) inaccessible cardinal, since any
weakly inaccessible cardinal is strongly inaccessible in
$L$, and so the issue about weakly or strongly inaccessible
is entirely irrelevant when it comes to consistency.

Second, let me note that set theorists are not generally
satisfied by claims of the sort "the only known proof uses
such-and-such," but rather they use the concepts of
consistency strength and equiconsistency, which allow for
precise claims to be proved about exactly which large
cardinals are required to prove which statements. The
situation is that for many mathematical assertions, we can
prove that any proof *must* use a certain type of large
cardinal or something just as strong, in the sense that the
consistency of the statement itself implies the consistency
of the large cardinal in question. In this way, we avoid
any problematic issue about knowledge concerning whether a better proof is
simply not yet discovered.

As a result, if inaccessible cardinals should be refuted,
then using the known results we immediately gain an
enormous number of positive theorems. So it isn't really a
case of losing theorems, but rather gaining.

**Theorem.** If inaccessible cardinals are inconsistent,
then (we can prove that) we can construct a non-Lebesgue
measurable set of reals without using the axiom of choice.

This follows from the fact that Solovay and Shelah have
proved that the possibility of constructing a non-Lebesgue
measurable set of reals (in the context of ZF+DC) without
using AC is exactly equivalent to the inconsistency of
inaccessible cardinals.

Most people believe that one must use AC in any Vitali-type
construction of a non-Lebesgue measurable set, and the
theorem above shows that this belief is provably equivalent
to the consistency of inaccessible cardinals. Perhaps many
mathematicians would find their confidence in the
consistency of inaccessible cardinals to increase upon
learning of this, and in this sense, this is also an answer to question 1.
In any case, many well-known set
theorists have emphasized enormous confidence in the
consistency of large cardinals, and have stated quite
explicitly that if inaccessible cardinals should become
known to be inconsistent, then we should expect further
inconsistency much lower in ZFC itself or in the low levels
of PA.

**Theorem.** If inaccessible cardinals are inconsistent
(and even merely if we can refute infinitely many Woodin
cardinals), then (we can prove that) there is a projective
set of reals $A\subset\mathbb{R}$ whose corresponding
two-person game of perfect information has no winning
strategy for either player. In other words, the infinitary
de Morgan law
$$\neg\forall n_0\exists n_1\forall n_2\exists n_3\cdots A(\vec
n)\iff\exists n_0\forall n_1\exists n_2\forall
n_3\cdots\neg A(\vec n)$$ will fail for some projective set
$A$.

The projective sets of reals are those reals that are
definable by a property involving quantification only over
real numbers and integers. The reason for the theorem is
that projective determinacy is equiconsistent over ZFC with
infinitely many Woodin cardinals, and so if we refute the
large cardinals in ZFC, then we similarly refute projective
determinacy.

**Theorem.** If inaccessible cardinals are inconsistent
(and even if merely measurable cardinals are inconsistent),
then (we can prove that) there is an analytic set (a
continuous image of a Borel set) that is not determined.

**Theorem.** If inaccessible cardinals are inconsistent,
then we can prove that the full set-theoretic universe is
very close to the constructible universe in the sense of
covering. In particular, $L$ computes the successors of
singular cardinals correctly.

This shocking conclusion follows in this case from Jensen's
covering lemma, since refuting inaccessible cardinals
implies a refutation of $0^\sharp$.

**Theorem.** If inaccessible cardinals are inconsistent,
then on no set is there a countably complete real-valued
measure measuring all subsets of the set and giving points no mass.

This is simply because any real-valued measurable cardinal
is measurable and hence inaccessible in an inner model.

**Theorem.** If inaccessible cardinals are inconsistent,
then (we can prove that) there are no uncountable
Grothendieck universes and the axiom of universes in
category theory is false.

An uncountable Grothendieck universe is exactly $H_\kappa$
for an inaccessible cardinal $\kappa$, and the axiom of
universes asserts that every set is in such a universe.

There are many more examples. (I invite any knowledgeable
person to edit the answer with additional examples.)